OCR A Level Chemistry A: Alkenes (H432) – Revision Notes

For full explanations and longer worked examples, read the Alkenes study guide first.

These notes condense section 4.1.3, Alkenes, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They cover outcomes 4.1.3(a) to (l) in Module 4, Core organic chemistry, which the specification assesses in Synthesis and analytical techniques (02) and Unified chemistry (03). H432 is the full A level. PAG7 links to the bromine test in outcome (f).

Course links: OCR chemistry hub, printable checklist, Alkenes practice questions and the free diagnostics.

4.1.3(a) and (b): Bonding and shape

Term	Meaning
Alkene	Unsaturated hydrocarbon with a C=C bond
σ-bond	Orbitals overlap directly between the two bonded atoms
π-bond	Sideways overlap of adjacent p-orbitals; electron density above and below the bonding carbons
Restricted rotation	The π-bond stops the carbons rotating about C=C
Shape at each C=C carbon	Trigonal planar, about 120°

Why 120°: each C=C carbon has three regions of electron density (the double bond counts as one). They repel equally to be as far apart as possible. You do not need hybridisation.

4.1.3(c) and (d): Stereoisomerism

Word	What it means
Stereoisomers	Identical structural formula; atoms arranged differently in space
E/Z isomerism	Stereoisomerism from restricted rotation about C=C, needing two different groups on each C=C carbon
Cis–trans isomerism	Special case of E/Z where two of the substituent groups on each C=C carbon are the same

Method in steps: does a molecule have E/Z isomers?

1. Find every C=C.
2. For each C=C carbon, write its two groups.
3. If either carbon has two identical groups (=CH₂, =CCl₂, =C(CH₃)₂), that C=C gives no isomers.
4. If both carbons have two different groups, that C=C can be E or Z.
5. With more than one such C=C, list the combinations and remove any that are identical because the molecule is symmetrical.

Method in steps: assigning E or Z by CIP rules

1. On each carbon separately, compare the atoms bonded directly to it. Higher atomic number wins.
2. Tie? Compare the atoms one bond further out, highest first.
3. Higher-priority groups on the same side: Z. Opposite sides: E.

Atomic numbers you will use most: I 53, Br 35, Cl 17, F 9, O 8, N 7, C 6, H 1.

Small reminder. 1-iodopropene, ICH=CHCH₃. Left carbon: I beats H. Right carbon: CH₃ beats H. Iodine and methyl on the same side gives the Z isomer, which is also cis because each carbon carries an H.

E = trans and Z = cis are only reliably equivalent when each C=C carbon has an H. With heavier groups, rank them by CIP every time. You will not be asked to rank multiple bonds inside R groups.

4.1.3(e): Reactivity

The π-bond has a relatively low bond enthalpy, so it breaks more easily than the σ-bond. Estimate: π ≈ (C=C) − (C–C) from mean bond enthalpies. The π electrons are exposed above and below the plane, which attracts electrophiles.

4.1.3(f): Addition reactions

Reagent	Condition	Product	Example
H₂	Ni catalyst, heat	Alkane	hept-2-ene → heptane
Cl₂ or Br₂	Room temperature	Dihaloalkane	oct-2-ene + Br₂ → 2,3-dibromooctane
HCl, HBr or HI	Room temperature	Haloalkane	dec-5-ene + HBr → 5-bromodecane
H₂O(g)	Acid catalyst, e.g. H₃PO₄	Alcohol	ethene + H₂O → ethanol

Bromine test (PAG7). Add bromine water and shake. Alkene present: orange to colourless. No C=C: stays orange.

4.1.3(g) and (h): Electrophile and mechanism

• Electrophile: an electron pair acceptor.
• Heterolytic fission: a bond breaks and one atom takes both electrons, giving a cation and an anion.

Method in steps: electrophilic addition with H–X

1. Draw the alkene and H–X with δ+ on H and δ− on X.
2. Arrow 1 starts at the C=C bond and ends at H.
3. Arrow 2 starts at the H–X bond and ends at X.
4. Draw the carbocation (+ on the carbon that did not gain H) and X⁻ with a lone pair.
5. Arrow 3 goes from the lone pair on X⁻ to the C⁺.
6. Draw the haloalkane.

With Br₂: the π electrons induce a dipole in Br–Br; the nearer Br is δ+. Same three arrows. The intermediate may be shown as a carbocation or a bromonium ion; the specification accepts either.

4.1.3(i): Markownikoff’s rule

Stability: tertiary > secondary > primary carbocation. H–X adding to an unsymmetrical alkene gives a major product from the more stable carbocation. No explanation of the stability order is needed.

Method in steps: predicting the major product

1. Number the two C=C carbons.
2. Put H on the first carbon and + on the second. Classify the carbocation.
3. Swap and classify again.
4. The more substituted carbocation leads to the major product; X ends up on that carbon.

Small reminder. 2-methylpent-2-ene, (CH₃)₂C=CHCH₂CH₃, with HI. + on C2 is tertiary; + on C3 is secondary. Major product: 2-iodo-2-methylpentane.

4.1.3(j): Addition polymers

Method in steps: monomer to repeat unit

1. Draw the monomer with all four groups around C=C.
2. Open the double bond to a single bond.
3. Add one extension bond on each side and put the unit in square brackets.

Method in steps: polymer to monomer

1. Find the shortest repeating section: two main-chain carbons.
2. Put back the C=C between them and keep their side groups.

4.1.3(k) and (l): Waste polymers

Approach	Benefit
Combustion for energy	Energy from waste is used, not buried in landfill
Organic feedstock	Polymers broken into small molecules for new plastics and chemicals; less crude oil used
Removing toxic products	HCl from burning PVC is neutralised before gases are released
Biodegradable polymers	Broken down by microorganisms; less persistent waste; often made from plant sources
Photodegradable polymers	Bonds break when light is absorbed; less persistent waste

Must-know distinctions

• σ vs π: σ overlaps directly between nuclei and allows rotation; π overlaps sideways and prevents it.
• E/Z vs cis–trans: every cis–trans pair is an E/Z pair, but not every E/Z pair can be named cis or trans.
• Cis vs Z: cis compares identical groups; Z compares higher-priority groups. They can disagree.
• Heterolytic vs homolytic: heterolytic gives ions (alkene addition); homolytic gives radicals (alkane substitution in 4.1.2).
• Biodegradable vs photodegradable: microorganisms versus light.

Quick self-test

Data: C = 12.0, H = 1.0, Br = 79.9; molar gas volume at RTP = 24.0 dm³ mol⁻¹.

1. Define the term electrophile.
2. State the bond angle around each carbon of a C=C bond and name the shape.
3. Which of hex-1-ene, hex-2-ene and 2-methylpent-2-ene has E/Z isomers?
4. FCH=CHI has F and I on opposite sides of the double bond. Is it E or Z?
5. State the reagent and catalyst that convert hept-2-ene into heptane.
6. What volume of H₂, at RTP, is used up when 3.36 g of but-1-ene is fully hydrogenated?
7. Calculate the mass of bromine that reacts with 4.20 g of hex-2-ene.
8. Give the major product of but-1-ene with HI and the type of carbocation it forms from.
9. Draw the repeat unit of the polymer made from but-1-ene.
10. A polymer section is –CH₂–CH(CH₂CH₂CH₃)–CH₂–CH(CH₂CH₂CH₃)–. Name the monomer.
11. Name the toxic gas formed when PVC is burned and say how it is removed.

Answers

1. An electron pair acceptor.
2. About 120°; trigonal planar.
3. Hex-2-ene only. Hex-1-ene has =CH₂; in 2-methylpent-2-ene, C2 carries two CH₃ groups.
4. E (F beats H, I beats H; they are on opposite sides). It is also trans.
5. Hydrogen, with a nickel catalyst.
6. Mr C₄H₈ = 56.0, so 3.36 g is 0.0600 mol, which takes up 0.0600 mol H₂: 0.0600 × 24.0 = 1.44 dm³.
7. Mr C₆H₁₂ = 84.0. Amount = 4.20 ÷ 84.0 = 0.0500 mol. Mr Br₂ = 159.8. Mass = 0.0500 × 159.8 = 7.99 g.
8. 2-iodobutane, from the secondary carbocation (the primary one gives the minor product, 1-iodobutane).
9. –[CH₂–CH(CH₂CH₃)]– with an extension bond on each side.
10. Pent-1-ene, CH₂=CHCH₂CH₂CH₃.
11. Hydrogen chloride; the waste gases are passed through a base, which neutralises it.

Where marks are usually lost

• Writing “the π-bond is between the two carbons” instead of above and below them.
• Defining an electrophile as “something that attacks a double bond” rather than an electron pair acceptor.
• Claiming E/Z isomers for an alkene with two identical groups on one C=C carbon.
• Using size or relative mass for CIP priority instead of atomic number.
• Naming cis–trans and E/Z as if they always agree.
• Starting a curly arrow at a carbon atom, or ending the first arrow on the C=C instead of the H.
• Missing the lone pair on the halide ion, or the dipole on H–X.
• Putting the halogen on the wrong carbon because the carbocations were not compared.
• Leaving C=C in a repeat unit, or giving a four-carbon repeat unit for a simple polymer.
• Saying “clear” for the bromine test result.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.1.3 Alkenes, learning outcomes (a) to (l). Published by OCR.